If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 3.8 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is \( -32 \mathrm{ft} / \mathrm{sec}^{2} \). [? ] feet
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Bonus Knowledge
To determine the maximum height the ball reached, we can use the kinematic equation for projectile motion: \[ h(t) = h_0 + v_0 t + \frac{1}{2} a t^2 \] where: - \( h(t) \) is the height at time \( t \), - \( h_0 \) is the initial height (4 feet), - \( v_0 \) is the initial velocity (which we will need to calculate later), - \( a \) is the acceleration due to gravity (-32 ft/s²), - \( t \) is the time in seconds. Given that the ball remains in the air for 3.8 seconds, it takes half of that time (1.9 seconds) to reach its peak height. To find the initial velocity, we use the time taken to reach the peak, where the final velocity is 0: \[ v_f = v_0 + a t \] \[ 0 = v_0 - 32(1.9) \] Thus: \[ v_0 = 32(1.9) = 60.8 \text{ ft/s} \] Now, let's find the maximum height by substituting \( t = 1.9 \) seconds back into the height formula: \[ h(1.9) = 4 + 60.8(1.9) + \frac{1}{2}(-32)(1.9^2) \] Calculating each term: \[ h(1.9) = 4 + 115.52 - 57.76 = 61.76 \text{ feet} \] So, rounded to the nearest foot, the maximum height of the ball is approximately \(\boxed{62}\) feet.
